Problem Description

Alice and Bob are playing a stone removal game with specific rules:

Game Setup:

• There is a pile containing n stones
• Alice always goes first

Game Rules:

• Alice's first move: She must remove exactly 10 stones
• Subsequent moves: Each player must remove exactly 1 fewer stone than what their opponent removed in the previous turn
  • For example, if Alice removes 10 stones, Bob must remove 9 stones
  • If Bob removes 9 stones, Alice must remove 8 stones, and so on

Winning Condition:

• The player who cannot make a valid move (doesn't have enough stones to remove) loses the game
• The other player wins

Task: Given a positive integer n representing the initial number of stones in the pile, determine if Alice can win the game assuming both players play optimally. Return true if Alice wins, false if Bob wins.

Example walkthrough: If n = 12:

• Alice removes 10 stones, leaving 2 stones
• Bob needs to remove 9 stones but only 2 remain
• Bob cannot make a valid move, so Alice wins
• Return true

If n = 21:

• Alice removes 10 stones, leaving 11 stones
• Bob removes 9 stones, leaving 2 stones
• Alice needs to remove 8 stones but only 2 remain
• Alice cannot make a valid move, so Bob wins
• Return false

Intuition

The key insight is that this game has a deterministic outcome - once we know the starting number of stones n, we can determine who will win because both players must follow strict rules about how many stones to remove.

Let's trace through what happens:

• Alice removes 10 stones
• Bob must remove 9 stones
• Alice must remove 8 stones
• Bob must remove 7 stones
• And so on...

The sequence of stones removed is: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, ...

Notice that each player alternates turns, and the number of stones they must remove decreases by 1 each time. The game continues until someone cannot make their required move.

The crucial observation is that the winner depends on how many valid moves can be made in total. If the total number of moves is odd, Alice wins (since she goes first). If it's even, Bob wins.

Why? Because:

• If there's 1 move possible: Alice makes it and wins
• If there are 2 moves possible: Alice makes the first, Bob makes the second, Bob wins
• If there are 3 moves possible: Alice → Bob → Alice, Alice wins
• And so on...

So our strategy is simple:

1. Simulate the game by subtracting stones in the sequence 10, 9, 8, 7, ... until we can't anymore
2. Count how many moves were made
3. If the count is odd, Alice wins; if even, Bob wins

We don't need to track whose turn it is explicitly - we just need to count the total number of valid moves that can be made with the given number of stones.

Learn more about Math patterns.

Solution Approach

Based on our intuition, we implement a simulation that counts the total number of valid moves in the game.

Implementation Steps:

1. Initialize variables:

   • x = 10: The current number of stones to be removed (starts at 10 for Alice's first turn)
   • k = 0: Counter for the total number of moves made
   • n: The initial pile of stones

2. Simulate the game with a while loop:

   while n >= x:
       n -= x      # Remove x stones from the pile
       x -= 1      # Next player must remove 1 fewer stone
       k += 1      # Increment the move counter

   The loop continues as long as there are enough stones (n >= x) to make the current required move.

3. Determine the winner:

   return k % 2 == 1

   • If k is odd, Alice wins (returns True)
   • If k is even, Bob wins (returns False)

Example Walkthrough:

For n = 21:

• Round 1: n = 21, x = 10 → Remove 10 stones, n = 11, x = 9, k = 1
• Round 2: n = 11, x = 9 → Remove 9 stones, n = 2, x = 8, k = 2
• Round 3: n = 2, x = 8 → Cannot remove 8 stones (only 2 left), loop exits
• Result: k = 2 (even) → Bob wins, return False

Time Complexity: O(√n) - In the worst case, we remove stones in the sequence 10 + 9 + 8 + ... + 1, which sums to approximately n. The number of terms in this sequence is roughly proportional to √n.

Space Complexity: O(1) - We only use a constant amount of extra space for variables x and k.

Example Walkthrough

Let's walk through a small example with n = 30 stones to illustrate the solution approach:

Initial Setup:

• n = 30 (stones in pile)
• x = 10 (stones Alice must remove)
• k = 0 (move counter)

Simulation:

Move 1 (Alice's turn):

• Check: Is n >= x? → Is 30 >= 10? → Yes, can make move
• Remove stones: n = 30 - 10 = 20
• Update for next move: x = 10 - 1 = 9 (Bob must remove 9)
• Increment counter: k = 1

Move 2 (Bob's turn):

• Check: Is n >= x? → Is 20 >= 9? → Yes, can make move
• Remove stones: n = 20 - 9 = 11
• Update for next move: x = 9 - 1 = 8 (Alice must remove 8)
• Increment counter: k = 2

Move 3 (Alice's turn):

• Check: Is n >= x? → Is 11 >= 8? → Yes, can make move
• Remove stones: n = 11 - 8 = 3
• Update for next move: x = 8 - 1 = 7 (Bob must remove 7)
• Increment counter: k = 3

Move 4 (Bob's turn):

• Check: Is n >= x? → Is 3 >= 7? → No, cannot make move
• Loop exits

Determine Winner:

• Total moves made: k = 3
• Is k odd? → 3 % 2 == 1 → Yes
• Since the total number of moves is odd, Alice wins!
• Return true

The game ended because Bob needed to remove 7 stones but only 3 remained in the pile. Bob couldn't make a valid move, so Alice wins. This matches our algorithm's prediction: with 3 total moves (odd number), Alice makes the last valid move and wins.

Solution Implementation

Time and Space Complexity

The time complexity is O(√n), where n is the number of stones. The loop starts with x = 10 and continues while n >= x, decreasing x by 1 in each iteration. The loop terminates when the remaining stones n are less than the current required stones x. Since we're subtracting decreasing values (10, 9, 8, ...) from n, the total sum of stones removed follows the pattern of triangular numbers. The sum 10 + 9 + 8 + ... + (10 - k + 1) approximately equals k * (21 - k) / 2. When this sum approaches n, we have k * (21 - k) / 2 ≈ n, which gives us k ≈ O(√n). Therefore, the number of iterations is proportional to √n.

The space complexity is O(1) as the algorithm only uses a constant amount of extra space for the variables x, k, and the modified n, regardless of the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Determining the Winner Based on Who Made the Last Move

The Pitfall: A common mistake is thinking that the player who makes the last valid move loses the game. This leads to incorrect logic like:

# WRONG interpretation
return total_turns % 2 == 0  # Thinking Alice loses if she makes the last move

Why This Happens: The problem statement says "The player who cannot make a valid move loses," which some might misinterpret as "the player who makes the last move loses."

The Correct Understanding:

• The player who cannot make a valid move loses
• The player who makes the last valid move wins
• If total_turns is odd, Alice made the last move and wins
• If total_turns is even, Bob made the last move and wins

2. Off-by-One Error in the Decrement Logic

The Pitfall: Decrementing stones_to_remove at the wrong point in the loop:

# WRONG: Decrementing before the removal
while n >= stones_to_remove:
    stones_to_remove -= 1  # Wrong position!
    n -= stones_to_remove
    total_turns += 1

Why This Matters:

• Alice's first move must remove exactly 10 stones, not 9
• Each subsequent move should be 1 less than the previous move
• Decrementing before removal would make Alice remove 9 stones first

The Fix: Always decrement after removing stones:

while n >= stones_to_remove:
    n -= stones_to_remove
    stones_to_remove -= 1  # Correct position
    total_turns += 1

3. Not Handling Edge Cases Properly

The Pitfall: Not considering what happens when n < 10:

# Might cause issues if not handled properly
if n < 10:
    # What should happen here?

The Solution: The while loop condition n >= stones_to_remove naturally handles this case:

• If n < 10, the loop never executes
• total_turns remains 0
• Alice cannot make her first move and loses immediately
• Returns False (correct behavior)

4. Using the Wrong Initial Value for stones_to_remove

The Pitfall: Starting with the wrong number:

stones_to_remove = 9  # WRONG: Should be 10
# or
stones_to_remove = 1  # WRONG: Misunderstanding the sequence

Why This Matters: The problem explicitly states Alice must remove exactly 10 stones on her first turn. Starting with any other value will produce incorrect results for all test cases.